A block is dragged on a smooth plane with the help of a rope which moves with a velocity v as shown in figure. The horizontal velocity of the block is :

Options

A. `v`
B. ` v sin theta `
C. v//sin theta`
D. `v//cos theta`

Answer

C

Explanation

Let ` AB = X, then ` AC= x cos theta`
:. ` (d(AC)/(dt) = d/(dt) (x cos theta) = (dx)/(dt) cos theta + x (- sin theta) (d theta)/(dt)`
But `(d(AC)/(dt) = 0 `, so ` 0= (dx)/(dt) cos theta + x ( - sin theta) ( d theat)/(dt)`
or ` v cos theta = x sin theta = x sin theta (d theta)/(dt)` or ` (dtheta)/(dt) = ( v cos theta)/(x sin theta)`
.
Also , ` CB = x sin theta`.
Velocity of block will be
` u = sin theta + x cos theta xx (v cos theta)/( x sin theta)`
` (v (sin^@ theta + cos^2 theta)/( sin theta) = v/(sin theta)`.